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Chapter 4: Reaction Stoichiometry • • • • • • Mole-to-mole ratios and calculations [4.2] Limiting Reactants and Yields [4.3] Molarity and stoichiometry [4.4] Solubility Rules [4.5] Types of Reactions [4.6, 8, 9] Writing reactions [4.7] CHEM& 141 F08 1 Reaction Stoichiometry • For all reactions, we need to do many calculations: – How much of each starting material is needed? – How much product is made? – What materials are left over? • These calculations follow methods we have already learned: g → mol → mol → g • A balanced equation is crucial to the success of problem-solving! – Coefficients provide mole-to-mole ratio! CHEM& 141 F08 2 1 Reaction Stoichiometry Steps for problem-solving: 1) Make sure you have a balanced equation. 2) Calculate molar masses of the compounds you are interested in. 3) Determine the mole-to-mole ratio from the coefficients in the balanced equation. 4) Set-up and solve the problem, watching units to guide the process. CHEM& 141 F08 3 Reaction Stoichiometry How much Fe will be produced from 152.6 g of carbon monoxide and an excess of iron (III) oxide? Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) 28.01 g/mol 55.85 g/mol g CO ! mol CO ! mol Fe ! g Fe 1 mol CO 2 mol Fe 55.85 g Fe 152.6 g CO x x x = 202.8 g Fe 28.01 g CO 3 mol CO 1 mol Fe Start with the mass amount given in problem statement. CHEM& 141 F08 Use the mol-to-mol ratio from the coefficients in the balanced equation Double-check that your units cancel, and you have the correct sig figs! 4 2 Reaction Stoichiometry How many grams of CO2 are produced in the combustion of 50.0 g of propane, C3H8? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) CHEM& 141 F08 5 Reaction Stoichiometry Determine the amount of oxygen produced when 0.549 g of KClO3 decomposes. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) CHEM& 141 F08 6 3 Limiting Reactants In all of the previous examples, the amount of only one starting material is given. What if you know about both? Example: In the summer, I love fresh tomato sandwiches. I like 3 slices of tomatoes per sandwich on two slices of bread. If I have 20 slices of bread and 24 slices of tomatoes, how many sandwiches can I make? What do I have left over, if anything? CHEM& 141 F08 7 Limiting Reactants - Demo Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) CHEM& 141 F08 0.6 g Mg 1.2 g Mg 2.4 g Mg 100 mL 1.0 M HCl 100 mL 1.0 M HCl 100 mL 1.0 M HCl 8 4 Limiting Reactants Steps for problem-solving: 1) Make sure you have a balanced equation. 2) Calculate the molar masses of all compounds that you are interested in. 3) From each starting material, determine the amount of the desired product that can be formed. 4) Compare the two product amounts; the lower amount is the amount that will be formed, and the reactant is the limiting reactant (also called limiting reagent). CHEM& 141 F08 9 Limiting Reactants Methanol (CH3OH) can be burned as an alternative to fossil fuels. For the following reaction, determine the amount of CO2 that is formed when 130.0 g of CH3OH reacts with 150.0 g of O2. 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) 32.04 g/mol 32.00 g/mol 44.01 g/mol CH 3OH: 130.0 g CH 3OH x O 2 : 150.0 g O 2 x 1 mol CH 3OH 32.04 g CH 3OH 1 mol O 2 32.00 g O 2 x 2 mol CO 2 3 mol O 2 x x 2 mol CO 2 2 mol CH 3OH 44.01 g CO 2 1 mol CO 2 x 44.01 g CO 2 1 mol CO 2 = 178.6 g CO 2 = 137.5 g CO 2 Less CO2 is produced from the oxygen, so oxygen is the limiting reactant and the theoretical yield is 137.5 g of CO2. CHEM& 141 F08 10 5 Limiting Reactants The compound IF5 can be prepared by the reaction of SF4 with I2O5. The only other product obtained is SO2. 5 SF4 + 2 I2O5 → 4 IF5 + 5 SO2 What is the maximum number of grams of IF5 that can be obtained from 40.0 g of SF4 and 40.0 g of I2O5? CHEM& 141 F08 11 Limiting Reactants Consider the following unbalanced reaction: TiO2 + C→ Ti + CO When 28.6 g of C is allowed to react with 88.2 g of TiO2, how many grams of Ti can be produced? Which reactant is the limiting reactant? Which reactant is in excess? How much of that reactant is leftover? CHEM& 141 F08 12 6 Limiting Reactants Consider the following unbalanced reaction: 1 TiO2 + 2 C → 1 Ti + 2 CO When 28.6 g of C is allowed to react with 88.2 g of TiO2, how many grams of Ti can be produced? Which reactant is the limiting reactant? Which reactant is in excess? How much of that reactant is leftover? 1 mol C 1 mol Ti 47.88 g Ti x x = 57.0 g Ti 12.011 g C 2 mol C 1 mol Ti 1 mol TiO 2 1 mol Ti 47.88 g Ti TiO 2 : 88.2 g TiO 2 x x x = 52.9 g Ti 79.87 g TiO 2 1 mol TiO 2 1 mol Ti C: 28.6 g C x TiO 2 is the limiting reagent, 52.9 g Ti can be produced from the reaction. C is in excess, to determine the amount leftover: 57.0 g Tifrom C - 52.9 g Tifrom TiO = 4.1 g excess Tifrom C 2 1 mol Ti 2 mol C 12.011 g C 4.1 g Ti x x x = 2.1 g C in excess 47.88 g Ti 1 mol Ti 1 mol C CHEM& 141 F08 13 Percent Yield • When we determine the amount of a product, we are calculating a theoretical yield for the reaction. • Unfortunately, no reaction is perfect! Percent Yield = CHEM& 141 F08 Actual Yield x 100% Theoretical Yield 14 7 Percent Yield Example: For the IF5 reaction, suppose only 47.68 g of IF5 were isolated. What is the percent yield for the reaction? Example: For the methanol combustion reaction, suppose only 75.0 g of carbon dioxide is formed. What is percent yield for the reaction? CHEM& 141 F08 15 Solutions - Definitions • Solute • Solvent • Solution • Electrolyte – Strong electrolytes – Weak electrolytes • Nonelectrolyte CHEM& 141 F08 16 8 Molarity and Solutions Many reactions are done in the solution phase, where the reactants are dissolves in a solvent. Molarity (M) = moles of solute L of solution Problem A 1.192 g sample of oxalic acid, H2C2O4, is placed in water to reach a total of 100.0 mL. What is the molar concentration of the resulting solution? CHEM& 141 F08 17 Molarity Problems 1.000 g of sulfuric acid is required for a reaction. How many mL of a 0.200 M of sulfuric acid is needed to obtain this? How do you prepare 500.0 mL of 0.25 M HCl solution from concentrated HCl, which is 12.4 M? CHEM& 141 F08 18 9 Atoms Mass Atoms Moles A Moles B Volume Mass Volume CHEM& 141 F08 19 Molarity and Stoichiometry An aluminum can will dissolve in an aqueous solution of strong base, according to the following reaction: 2 Al(s) + 2 KOH(aq) + 6 H2O(l) 2 KAl(OH)4(aq) + 3 H2(g) 2.05 g of Al are placed in a beaker with 185 mL of 1.35 M KOH. What is the limiting reagent? How much (in g) KAl(OH)4 is produced? CHEM& 141 F08 20 10 Writing and Predicting Equations • We will focus on aqueous reactions, or reactions where water is the solvent. – Biological relevance – Environmental relevance • What causes a reaction to “go” or proceed? – A reaction needs a “driving force”! • A solid forms, called a precipitate (ppt) • A gas forms, ↑ • Water forms • Electrons are exchanged CHEM& 141 F08 21 Writing and Predicting Equations Reactions can be classified/grouped in two different ways: Chemical • Precipitation Pattern • – 2 ionic compounds react to form 2 new ionic compounds (or water) – A solid forms • Acid/Base – Water forms – Some gas-forming • Reduction/Oxidation, or “Redox” – Electrons exchange – Elements ⇔ compounds – Some gas-forming CHEM& 141 F08 Double Displacement • Single Displacement – 1 ionic compound + 1 element yield a new ionic compound and a new element • Combination • Decomposition – 2 or more reactants yield one product – One reactant yields 2 or more products 22 11 Writing Reactions • For reactions to proceed in water, the reactants should be soluble - they form a homogeneous solution. – Reactions proceed faster if this is the case! • How do we determine this? – Type of compound • Ionic - can be soluble or insoluble, based on rules • Molecular - often insoluble – Electrolyte behavior • Does the solution conduct electricity? CHEM& 141 F08 23 Ionic Compounds - Solubility Definitions (for ionic compounds mixing in H2O): Soluble - K The molecule breaks apart and individual ions “roam” independently. SO4 K H2O K+1 SO4 -2 K+1 K2SO4 → 2 K+ + SO4-2 Insoluble - The molecule stays together- ions stay joined. Ag CHEM& 141 F08 SO4 Ag H2O Ag SO4 Ag 24 12 Solubility Rules Ion(s) Li+, Na+, K+, NH4+ C2H3O2-, NO3Cl-, Br-, ISO42- Rule Group IA and ammonium salts are all soluble. Acetates and nitrates are soluble. Most halides are soluble. Most sulfates are soluble. CO32- Most carbonates are insoluble. 3- Most phosphates are insoluble. PO4 2- S Most sulfides are insoluble. - OH Most hydroxides are insoluble. Exceptions none none Ag+, Hg22+, Pb2+ CaSO4, SrSO4, BaSO4, Ag2SO4, Hg2SO4, PbSO4 Group IA and ammonium carbonates Group IA and ammonium phosphates Group IA and ammonium sulfides and CaS, SrS, BaS Group IA and ammonium hydroxides, Ca(OH)2, Sr(OH)2, Ba(OH)2 A similar table may be found in your text (Table 4.1). CHEM& 141 F08 25 Predicting Solubility • General Rules: – Look at cation first - they are the overriding predictor of solubility! • All compounds containing group I (including H+) and NH4 + ions are soluble - no exception! – If the cation does not make it soluble, then look at the anion. Ex: Determine if the following compounds are soluble or insoluble. If soluble, split into ions. CHEM& 141 F08 (NH4)2CO3 CuSO4 AgBr LiOH BaSO4 Cobalt (III) Sulfide Mercury (I) chloride Sodium phosphate 26 13 Writing Reactions - Double Displacement Most precipitation and acid/base reactions are double displacement reactions. How do we write these reactions? 1. Split the reactants into individual ions. 2. Swap the cations to form new compounds these are your products. 3. Write the new compounds with balanced formulas. 4. Determine the phases of the new products - are the new compounds soluble or insoluble? 5. Balance the overall equation. CHEM& 141 F08 27 Writing Reactions - Double Displacement Write a balanced equation with the following starting materials: NaNO3(aq) + CaCl2(aq) H2SO4(aq) + LiOH(aq) CHEM& 141 F08 28 14 Writing Reactions - Double Displacement Write a balanced equation with the following starting materials: Copper (II) nitrate and ammonium sulfide Hydrochloric acid and silver hydroxide CHEM& 141 F08 29 Redox Reactions - Definitions Most redox reactions are single displacement reactions. How do we recognize a redox reaction? Oxidation: Reduction: Oxidizing agent: Reducing agent: CHEM& 141 F08 30 15 Assigning Oxidation States • In free elements, each atom has an oxidation state of zero. • For single-atom ions, the oxidation state equals the charge of the atom. **Predict off periodic table!! • Hydrogen is either +1 or –1, depending on the atom it is bonded to. • Oxygen is typically –2, unless bonded to itself (e.g. H2O2) • The sum of the oxidation states must equal the charge of the molecule or ion. CHEM& 141 F08 31 Assigning Oxidation States Problem: Determine the oxidation states for each atom in the following compounds: CH3OH NaBH4 Perchlorate ion FeBr3 CHEM& 141 F08 32 16 Single Displacement Reactions How do we write these SD reactions? 1. Split the ionic compound into individual ions. 2. Take the elemental reactant and turn it into a cation of appropriate charge. 3. Swap the cations to form a new compound - this will be one of your products. 4. Take the remaining cation and convert it to an element this will be the other product. 5. Write the new compounds with balanced formulas. 6. Determine the phases of the new products - are the new compounds soluble or insoluble? 7. Balance the overall equation. CHEM& 141 F08 33 Writing Reactions - Single Displacement Write a balanced equation with the following starting materials: Ca(s) + HCl(aq) Cu(s) + Ag2SO4(aq) CHEM& 141 F08 34 17 Writing Reactions - Single Displacement Write a balanced equation with the following starting materials: Na(s) + H2O(l) Lead and hydroiodic acid to make a lead (II) compound CHEM& 141 F08 35 Net Ionic Equations We often “translate” chemical equations into net ionic equations (or reactions). These reactions strip out the ions that do not participate in the chemical process described by the reaction. Example: Molecular Equation CaCl2(aq) + Li2 SO4(aq) → CaSO4(s) + 2 LiCl(aq) Ionic Equation Ca+2(aq) + 2 Cl-(aq) + 2 Li+(aq) + SO4-2(aq) → CaSO4(s) + 2 Li+(aq) + 2 Cl-(aq) Net Ionic Equation Ca+2(aq) + SO4-2(aq) → CaSO4(s) CHEM& 141 F08 36 18 Net Ionic Equations Do the following steps to generate a net ionic equation: 1) You must begin with a balanced molecular equation, with all phase labels listed. 2) For each and only (aq) species, break it up into individual ions (keep polyatomic ion units together). 3) Do not break up any (s), (l), or (g). This generates the ionic equation. 4) Inspect the ionic equation - for each species that appears in the identical form on both sides of the equation, cross those out. Reconcile any species that appear in differing amounts on both sides. 5) Rewrite only the species that were not crossed out in a new reaction - the net ionic equation! CHEM& 141 F08 37 Net Ionic Equations Write net ionic equations for the following balanced equations. 2 AgNO3(aq) + MgCl2(aq) → 2 AgCl(s) + Mg(NO3)2(aq) 2 Ag+(aq) + 2 NO3-(aq) + Mg+2(aq) + 2 Cl-(aq) → 2 AgCl(s) + Mg+2(aq) + 2 NO3-(aq) 2 Ag+(aq) + 2 Cl-(aq) → 2 AgCl(s) H2SO4(aq) + Zn(s) → H2(g) + ZnSO4(aq) 2 NaOH(aq) + Pb(C2H3O2)2(aq) → Pb(OH)2(s) + 2 NaC2H3O2(aq) CHEM& 141 F08 38 19 Some Special Reactions Gas-Forming Reactions fall under either redox or acid/base reactions. Redox: Metal + acid → Hydrogen gas + ionic compound H2SO4(aq) + Zn(s) → H2(g) + ZnSO4(aq) 2 HClO4(aq) + Pt(s) → H2(g) + Pt(ClO4)2(aq) Acid/Base: Reactions that generate H2CO3 decompose further to CO2 and H2O CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2CO3(aq) → H2O(l) + CO2(g) CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) K2CO3(s) + 2 HBr(aq) → 2 KBr(aq) + H2O(l) + CO2(g) CHEM& 141 F08 39 Putting it all together Perform the following: 1) Write a complete and balanced chemical equation that describes the reaction between aluminum metal and sulfuric acid. 2) Determine how much 0.125 M sulfuric acid is needed (in mL) to react with 22.5 g of aluminum. 3) Determine how much of the elemental product is liberated in that reaction as well. CHEM& 141 F08 40 20